What is the integral of 1/((4x+1)^{10)} ?

The integral of 1/((4x+1)^{10)} is -1/(36(4x+1)^9)+C

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integral of 1/((4x+1)^{10)}

\int \frac{1}{( 4 x + 1 ) ^{10}} d x

- \frac{1}{36 ( 4 x + 1 ) ^{9}} + C

Solution steps

\int \frac{1}{( 4 x + 1 ) ^{10}} d x

Apply u-substitution

: \int \frac{1}{4 u ^{10}} d u

= \int \frac{1}{4 u ^{10}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{1}{u ^{10}} d u

Apply the Power Rule

: - \frac{1}{9 u ^{9}}

= \frac{1}{4} (- \frac{1}{9 u ^{9}})

Substitute back

u = 4 x + 1

= \frac{1}{4} (- \frac{1}{9 ( 4 x + 1 ) ^{9}})

Simplify

\frac{1}{4} (- \frac{1}{9 ( 4 x + 1 ) ^{9}}) : - \frac{1}{36 ( 4 x + 1 ) ^{9}}

= - \frac{1}{36 ( 4 x + 1 ) ^{9}}

Add a constant to the solution

= - \frac{1}{36 ( 4 x + 1 ) ^{9}} + C

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What is the integral of 1/((4x+1)^{10)} ?
The integral of 1/((4x+1)^{10)} is -1/(36(4x+1)^9)+C